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3 years ago

What happens when any ray of light strikes a concave mirror, after passing through its focal point?

Physics

2 answers:

nekit [7.7K]3 years ago

6 0

Answer:

Option (D)

Explanation:

The rules to draw the ray diagrams for concave mirror are given below:

1. When a ray of light incident on the pole of concave mirror at an angle of incidence, then after reflection, the angle of incidence is equal to the angle of reflection.

2. When incident ray of light passes through the centre of curvature of the mirror, then the reflected ray retraces its path.

3. When the incident ray of light passing through the focus of the concave mirror, then after reflection the reflected ray becomes parallel to the principal axis and vice versa.

baherus [9]3 years ago

5 0

D) The reflected ray travels parallel to the principal axis

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Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

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3 years ago

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3 years ago

A 10.0 Ω lightbulb is connected to a 12.0 V battery. (a) What current flows through the bulb? (b) What is the power of the bulb?

andreev551 [17]

The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

<h3>What current flows through the bulb as well as the power of the bulb?</h3>

From ohm's law; V = I × R

Where V is the voltage, I is the current and R is the resistance.

Also, Power is expressed as; P = V × I

Where V is voltage and I is current.

Given that;

Resistance R = 10.0 ohms
Voltage V = 12.0V
Current I = ?
Power P = ?

First, we determine the current flow through the bulb.

V = I × R

12.0V = I × 10.0 ohms

I = 12.0 ÷ 10.0

I = 1.2A

Next, we determine the power of the bulb.

P = V × I

P = 12.0V × 1.2A

P = 14.4 Watts

Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

Learn more about Ohm's law here: brainly.com/question/12948166

#SPJ1

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2 years ago